Use polynomial long division to rewrite the following fraction in the form $q(x)+\frac{r(x)}{d(x)}$, where $d(x)$ is the denominator of the original fraction, $q(x)$ is the quotient, and $r(x)$ is the remainder. \[ \frac{4 x^{3}-4 x^{2}+4 x+12}{2 x+2} \]

Step 1 ：We have the fraction \(\frac{4 x^{3}-4 x^{2}+4 x+12}{2 x+2}\). We can rewrite this as a polynomial division problem: \(4 x^{3}-4 x^{2}+4 x+12\) divided by \(2 x+2\).

Step 2 ：We can simplify the divisor \(2 x+2\) to \(x+1\) by factoring out 2.

Step 3 ：Now we perform polynomial long division. The first term of the quotient \(q(x)\) is \(4x^{2}\), because \(4x^{3}\) divided by \(x\) is \(4x^{2}\).

Step 4 ：Subtract \(4x^{2}(x+1)\) from \(4 x^{3}-4 x^{2}\) to get the new dividend \(-4x^{2}+4x+12\).

Step 5 ：The next term of the quotient is \(-4x\), because \(-4x^{2}\) divided by \(x\) is \(-4x\).

Step 6 ：Subtract \(-4x(x+1)\) from \(-4x^{2}+4x+12\) to get the new dividend \(4x+12\).

Step 7 ：The next term of the quotient is \(4\), because \(4x\) divided by \(x\) is \(4\).

Step 8 ：Subtract \(4(x+1)\) from \(4x+12\) to get the new dividend \(8\).

Step 9 ：Since the degree of the new dividend \(8\) is less than the degree of the divisor \(x+1\), we stop here. The remainder \(r(x)\) is \(8\).

Step 10 ：So, the original fraction can be rewritten as \(q(x)+\frac{r(x)}{d(x)}\), which is \(4x^{2}-4x+4+\frac{8}{x+1}\).

Step 11 ：Therefore, the final answer is \(\boxed{4x^{2}-4x+4+\frac{8}{x+1}}\).